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2 years ago

How can algebra help me with monthly finances and bill payments?

1 answer:

4 0

Personally, I don't see how algebra could help. Algebra is an advance form of mathematics. For monthly finances, you would need to do the following:
1. Figure out your how much you make monthly ( I would round to the nearest 5)
2. Decide what needs to be paid (i.e. bills, groceries, etc)
3. Figure in luxury expenses ( going to the movies, dates, etc.) and then just figure out what costs you can cut to leave you with enough money to live on. No complicated equations needed

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Katarina [22]

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hablas verja ijueputa mamaverja

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2 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

1 year ago

the parabola y=x^2 is shifted up by 2 units and to the right by 3 units. what is the equation of the new porabola

Juli2301 [7.4K]

Up 2 units = y shifted up by 2:
y = x^2 + 2
to the right 3 units = x shifted right (+) by 3:
y + 3 = x^2 +2
new equation:
y = x^2 - 1

6 0

2 years ago

Write the number 164 in expanded form

Lapatulllka [165]

If you were to write 164 in expanded form, it would be 100 + 60 + 4 = 164
Hope this helps and can I get brainliest answer bc i need it to rank up!! :D

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2 years ago

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Brainlist Tell me a joke, and you get brainlist

Lady_Fox [76]

Answer:

Why did the chicken cross the road?

To get to the other side

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2 years ago

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